Integrand size = 24, antiderivative size = 82 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=-\frac {2^{\frac {1}{2}+m} a c \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m}}{3 f} \] Output:
-1/3*2^(1/2+m)*a*c*cos(f*x+e)^3*hypergeom([3/2, 1/2-m],[5/2],1/2-1/2*sin(f *x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^(-1+m)/f
Result contains complex when optimal does not.
Time = 2.16 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.60 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=-\frac {2^{-m} c \left (i+e^{i (e+f x)}\right ) \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^m \left (-\left ((-1+m) m \operatorname {Hypergeometric2F1}\left (1,m,-m,i e^{i (e+f x)}\right )\right )+e^{i (e+f x)} (1+m) \left (-2 i (-1+m) \operatorname {Hypergeometric2F1}\left (1,1+m,1-m,i e^{i (e+f x)}\right )+e^{i (e+f x)} m \operatorname {Hypergeometric2F1}\left (1,2+m,2-m,i e^{i (e+f x)}\right )\right )\right ) (-1+\sin (e+f x))}{\left (-i+e^{i (e+f x)}\right )^2 f (-1+m) m (1+m)} \] Input:
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]),x]
Output:
-((c*(I + E^(I*(e + f*x)))*(((-I)*a*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x )))^m*(-((-1 + m)*m*Hypergeometric2F1[1, m, -m, I*E^(I*(e + f*x))]) + E^(I *(e + f*x))*(1 + m)*((-2*I)*(-1 + m)*Hypergeometric2F1[1, 1 + m, 1 - m, I* E^(I*(e + f*x))] + E^(I*(e + f*x))*m*Hypergeometric2F1[1, 2 + m, 2 - m, I* E^(I*(e + f*x))]))*(-1 + Sin[e + f*x]))/(2^m*(-I + E^(I*(e + f*x)))^2*f*(- 1 + m)*m*(1 + m)))
Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3215, 3042, 3168, 80, 79}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (c-c \sin (e+f x)) (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c-c \sin (e+f x)) (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle a c \int \cos ^2(e+f x) (\sin (e+f x) a+a)^{m-1}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a c \int \cos (e+f x)^2 (\sin (e+f x) a+a)^{m-1}dx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^3 c \cos ^3(e+f x) \int \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m-\frac {1}{2}}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^3 c 2^{m-\frac {1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-2} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}} \sqrt {a-a \sin (e+f x)}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {a^2 c 2^{m+\frac {1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x]),x]
Output:
-1/3*(2^(1/2 + m)*a^2*c*Cos[e + f*x]^3*Hypergeometric2F1[3/2, 1/2 - m, 5/2 , (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^ (-2 + m))/f
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin [e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))) Subst[Int[(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
\[\int \left (a +\sin \left (f x +e \right ) a \right )^{m} \left (c -c \sin \left (f x +e \right )\right )d x\]
Input:
int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e)),x)
Output:
int((a+sin(f*x+e)*a)^m*(c-c*sin(f*x+e)),x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="fricas")
Output:
integral(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=- c \left (\int \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- \left (a \sin {\left (e + f x \right )} + a\right )^{m}\right )\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e)),x)
Output:
-c*(Integral((a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral(-(a*sin(e + f*x) + a)**m, x))
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="maxima")
Output:
-integrate((c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int { -{\left (c \sin \left (f x + e\right ) - c\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="giac")
Output:
integrate(-(c*sin(f*x + e) - c)*(a*sin(f*x + e) + a)^m, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (c-c\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:
int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)),x)
Output:
int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=c \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m}d x -\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right )\right ) \] Input:
int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x)
Output:
c*(int((sin(e + f*x)*a + a)**m,x) - int((sin(e + f*x)*a + a)**m*sin(e + f* x),x))